How do you solve #4(u+1)^2=18#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Tony B Jun 22, 2016 #u= -1+-(3sqrt(2))/2# Explanation: #4(u+1)^2=18larr# Divide both sides by 4 #(u+1)^2=18/4 larr# Square root both sides #u+1=+-sqrt(18/4)larr# Subtract 1 from both sides #u=+-sqrt(18/4)-1larr# simplify the square root #u=-1+-sqrt(2xx3^2)/sqrt(2^2)larr# #u= -1+-(3sqrt(2))/2# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 2012 views around the world You can reuse this answer Creative Commons License